A-Level Maths: Worked Examples Explained | A-Level 数学：典型例题详解

📚 A-Level Maths: Worked Examples Explained | A-Level 数学：典型例题详解

Mastering A-Level Mathematics requires more than memorising formulas – it demands a deep understanding of how theory is applied in exam-style problems. In this article, we work through carefully selected examples from Pure Maths, Mechanics and Statistics, showing every logical step and common pitfalls. Each worked example is first explained in English, followed by a Chinese translation, so you can reinforce both your mathematical reasoning and bilingual terminology.

掌握 A-Level 数学不仅仅需要记忆公式，更需要对理论如何在考试题型中应用有深刻理解。本文精选了纯数、力学和统计的典型例题，逐一展示解题逻辑步骤与常见易错点。每道例题先用英文解释，再配以中文翻译，帮助你在巩固数学推理的同时掌握双语专业术语。

1. Algebraic Manipulation & Quadratic Equations | 代数运算与二次方程

Solve the quadratic equation 3x² − 5x − 2 = 0.

解二次方程 3x² − 5x − 2 = 0。

We try to factorise: look for two numbers that multiply to 3×(−2) = −6 and add to −5. The pair −6 and +1 works, so we split the middle term: 3x² − 6x + x − 2 = 0. Group: 3x(x − 2) + 1(x − 2) = 0 ⇒ (3x + 1)(x − 2) = 0. Thus x = −⅓ or x = 2.

尝试因式分解：寻找乘积为 3×(−2)=−6 且和为 −5 的两个数。−6 和 +1 符合要求，于是拆分中项：3x² − 6x + x − 2 = 0。分组：3x(x − 2) + 1(x − 2) = 0 ⇒ (3x + 1)(x − 2) = 0。因此 x = −⅓ 或 x = 2。

Always verify by expansion: (3x+1)(x−2) = 3x² − 6x + x − 2 = 3x² − 5x − 2, which matches.

始终用展开验证：(3x+1)(x−2) = 3x² − 6x + x − 2 = 3x² − 5x − 2，与左边一致。

Sometimes you must simplify before solving. For example, simplify (2x³ − 8x)/(2x) before differentiating. Divide each term by 2x: x² − 4. This avoids quotient rule mistakes.

有时解题前需先化简。例如在求导前将 (2x³ − 8x)/(2x) 化简，每一项除以 2x 得 x² − 4，从而避免商法则出错。

2. Differentiation & Tangents | 微分与切线方程

Given f(x) = 4x³ − 6x² + 2x − 7, find f'(x) and the equation of the tangent at x = 2.

已知 f(x) = 4x³ − 6x² + 2x − 7，求 f'(x) 以及当 x=2 时的切线方程。

Differentiating term by term: f'(x) = 12x² − 12x + 2.

逐项求导：f'(x) = 12x² − 12x + 2。

f'(x) = 12x² − 12x + 2

At x = 2, the gradient m = f'(2) = 12(4) − 12(2) + 2 = 48 − 24 + 2 = 26. The y‑coordinate is f(2) = 4(8) − 6(4) + 4 − 7 = 32 − 24 + 4 − 7 = 5. So we have the point (2,5) and m = 26.

当 x=2 时，斜率 m = f'(2) = 12(4) − 12(2) + 2 = 48 − 24 + 2 = 26。y 坐标为 f(2) = 4(8) − 6(4) + 4 − 7 = 32 − 24 + 4 − 7 = 5。得到点 (2,5) 和斜率 26。

Tangent equation: y − 5 = 26(x − 2) ⇒ y = 26x − 47. You should check it satisfies the original curve at x = 2.

切线方程：y − 5 = 26(x − 2) ⇒ y = 26x − 47。应验证 x=2 时与曲线一致。

Common error: forgetting to find the y‑coordinate first, or misapplying the power rule for negative exponents.

常见错误：忘记先求 y 坐标，或对负指数使用幂法则时出错。

3. Integration by Substitution | 换元积分法

Evaluate ∫ (2x + 1)√(x² + x) dx using the substitution u = x² + x.

利用换元 u = x² + x 计算 ∫ (2x + 1)√(x² + x) dx。

Set u = x² + x, so du/dx = 2x + 1. Thus du = (2x + 1) dx. The integral becomes ∫ √u du = ∫ u^½ du = (2/3) u^3/2 + C. Substituting back gives (2/3)(x² + x)^3/2 + C.

设 u = x² + x，则 du/dx = 2x + 1，从而 du = (2x + 1) dx。原积分化为 ∫ √u du = ∫ u^½ du = (2/3) u^3/2 + C。换回原变量得 (2/3)(x² + x)^3/2 + C。

∫ (2x + 1)√(x² + x) dx = ⅔ (x² + x)^3/2 + C

Always check your substitution yields a derivative present in the integrand; here du exactly matches (2x+1)dx. If the factor differs by a constant, adjust accordingly.

需始终检查换元后的导数是否在被积函数中出现；此处 du 恰好与 (2x+1)dx 匹配。若相差一个常数倍，需作相应调整。

For definite integrals, remember to change the limits: ∫_x=0¹ (2x+1)√(x²+x) dx, u(0)=0, u(1)=2, then evaluate from 0 to 2.

对于定积分，注意转换积分限：∫_x=0¹ (2x+1)√(x²+x) dx 中 u(0)=0, u(1)=2，然后从 0 到 2 计算。

4. Trigonometric Equations | 三角方程求解

Solve 2 sin²θ − sin θ − 1 = 0 for 0° ≤ θ ≤ 360°.

在 0° ≤ θ ≤ 360° 范围内解方程 2 sin²θ − sin θ − 1 = 0。

Treat it as a quadratic in sin θ: let y = sin θ ⇒ 2y² − y − 1 = 0. Factorising gives (2y + 1)(y − 1) = 0 ⇒ y = −½ or y = 1.

视作关于 sin θ 的二次方程：设 y = sin θ ⇒ 2y² − y − 1 = 0。因式分解得 (2y + 1)(y − 1) = 0 ⇒ y = −½ 或 y = 1。

For sin θ = 1, θ = 90°. For sin θ = −½, the reference angle is 30°, and since sine is negative in the third and fourth quadrants, θ = 180°+30° = 210° and θ = 360°−30° = 330°. Hence the solution set is {90°, 210°, 330°}.

对于 sin θ = 1，θ = 90°。对于 sin θ = −½，参考角为 30°，正弦在第三、四象限为负，故 θ = 180°+30° = 210° 以及 θ = 360°−30° = 330°。因此解集为 {90°, 210°, 330°}。

Always give answers within the specified range and check for extraneous solutions when squaring both sides.

始终给出指定范围内的答案，并在两边平方时检查增根。

5. Vectors & 3D Geometry | 向量与三维几何

Given points A(1, 2, −3) and B(4, 0, 5), find vector AB, its magnitude, and the angle between AB and AC where C has coordinates (3, 1, −2).

已知点 A(1, 2, −3) 和 B(4, 0, 5)，求向量 AB、其模长以及 AB 与 AC 的夹角，其中 C(3, 1, −2)。

AB = (4−1, 0−2, 5−(−3)) = (3, −2, 8). Magnitude |AB| = √(3² + (−2)² + 8²) = √(9+4+64) = √77.

AB = (3, −2, 8)。模长 |AB| = √(9 + 4 + 64) = √77。

AC = (3−1, 1−2, −2−(−3)) = (2, −1, 1). Its magnitude is |AC| = √(4+1+1) = √6. Dot product AB·AC = 3×2 + (−2)×(−1) + 8×1 = 6 + 2 + 8 = 16.

AC = (2, −1, 1)，模长 √6。点积 AB·AC = 16。

Using cos θ = (AB·AC) / (|AB| |AC|) = 16 / (√77 × √6). Hence θ = arccos(16/√462) ≈ 41.2°. Many candidates forget to take the dot product of the correct vectors.

由 cos θ = 16/(√77·√6) 得 θ ≈ 41.2°。许多考生容易点乘错误的向量对。

6. Arithmetic & Geometric Sequences | 等差与等比数列

In an arithmetic sequence the 3rd term is 10 and the 7th term is 22. Find the first term a and common difference d, then the sum of the first 20 terms.

某等差数列第 3 项为 10，第 7 项为 22。求首项 a 与公差 d，并计算前 20 项的和。

Using uₙ = a + (n−1)d: (1) a + 2d = 10, (2) a + 6d = 22. Subtract (1) from (2): 4d = 12 ⇒ d = 3. Then a = 10 − 2×3 = 4.

由通项公式得方程组，解出 d = 3，a = 4。

Sum S₂₀ = (n/2)[2a + (n−1)d] = 10[8 + 19×3] = 10[8+57] = 650.

求和 S₂₀ = (20/2)[2×4 + 19×3] = 10×65 = 650。

For geometric sequences you would use arⁿ⁻¹; be careful with the sign of the ratio r.

等比数列则用 arⁿ⁻¹，需注意公比 r 的正负。

7. Probability & Normal Distribution | 概率与正态分布

The random variable X ~ N(50, 4²). Find P(X < 55) and P(48 < X < 52).

随机变量 X 服从正态分布 N(50, 4²)。求 P(X < 55) 与 P(48 < X < 52)。

Standardise: Z = (X − μ)/σ. For X = 55, z = (55−50)/4 = 1.25. Using tables, Φ(1.25) = 0.8944, so P(X < 55) = 0.8944.

标准化：X=55 对应 z=1.25。查表得 Φ(1.25) = 0.8944，故 P(X < 55) = 0.8944。

For 48 < X < 52: z₁ = (48−50)/4 = −0.5, z₂ = 0.5. P = Φ(0.5) − Φ(−0.5) = 0.6915 − 0.3085 = 0.3830.

区间概率：z₁=−0.5, z₂=0.5，概率为 0.6915−0.3085 = 0.3830。

Always sketch the normal curve to confirm the area you need. Misreading Φ(–z) as 1−Φ(z) is a typical error.

应绘制正态曲线确认所需区域。误将 Φ(–z) 当作 1−Φ(z) 是典型错误。

8. Kinematics with Constant Acceleration | 匀加速运动学

A car accelerates uniformly from 5 m s⁻¹ to 25 m s⁻¹ in 8 seconds. Find the acceleration and the distance travelled during this period.

一辆汽车从 5 m s⁻¹ 匀加速至 25 m s⁻¹，用时 8 秒。求加速度及这段时间内的位移。

Using v = u + at: 25 = 5 + a×8 ⇒ 8a = 20 ⇒ a = 2.5 m s⁻².

用 v = u + at 得加速度 a = 2.5 m s⁻²。

Distance s = ut + ½at² = 5×8 + ½×2.5×64 = 40 + 80 = 120 m. Alternatively, average velocity (5+25)/2 × 8 = 15×8 = 120 m.

位移 s = 40 + 80 = 120 m。也可用平均速度法验证。

Beware of mixing up initial and final velocities; always assign u and v clearly. In vertical motion use g = 9.8 m s⁻² with sign conventions.

注意勿混淆初、末速度；竖直运动时需用 g = 9.8 m s⁻² 并遵循正负号约定。

9. Forces & Newton’s Laws | 力与牛顿定律

A block of mass 2 kg rests on a rough plane inclined at 30° to the horizontal. The coefficient of friction is 0.4. The block is released from rest. Determine its acceleration down the plane.

一个 2 kg 的物块静止在粗糙斜面上，斜面倾角 30°，摩擦系数 0.4。物块由静止释放，求其沿斜面向下的加速度。

Resolve weight: mg sin 30° = 2×9.8×0.5 = 9.8 N down the plane. Normal reaction R = mg cos 30° = 2×9.8×(√3/2) ≈ 16.97 N. Maximum friction F = μR = 0.4 × 16.97 ≈ 6.79 N.

分解重力：下滑分力 9.8 N。法向反力约 16.97 N，最大静摩擦 ≈ 6.79 N。

Net force down the plane = 9.8 − 6.79 = 3.01 N. Using F = ma, a = 3.01 / 2 ≈ 1.505 m s⁻².

沿斜面合力 3.01 N，加速度 a ≈ 1.505 m s⁻²。

Remember to check whether the component of weight overcomes friction; if not, acceleration is zero. Also, friction always opposes motion.

务必检验重力分量是否超过摩擦力，若未超过则加速度为零；摩擦始终与运动趋势方向相反。

10. Exam Strategy & Common Pitfalls | 考场策略与常见失分点

A recurring mistake is dropping the constant of integration ‘+C’ in indefinite integrals – examiners explicitly penalise this. Always write it, and adjust for definite integrals.

常见的失误是在不定积分中遗漏积分常数“+C”，阅卷中会明确扣分。必须写下，而定积分无需常数。

Another trap is confusing sin²θ + cos²θ ≡ 1 with double‑angle identities; labelling each identity correctly helps avoid substitution errors.

另一陷阱是混淆 sin²θ+cos²θ≡1 与倍角公式；清晰标注各恒等式可减少代入错误。

When using the product or chain rule, write the intermediate derivative steps to minimise sign errors. For mechanics, draw a clear free‑body diagram and mark all forces before writing equations.

使用乘积法则或链式法则时，写出中间导数步骤可减少符号错误。解力学题时，先画清晰的受力分析图并标出所有力，再列方程。

Finally, manage time wisely: tackle the longer, more demanding questions first while your mind is fresh. Always check units and state the final answer to the appropriate degree of accuracy.

最后，合理分配时间：趁头脑清醒时先攻克分值高、难度大的题目。务必检查单位并按合适的精度给出最终答案。

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A-Level Maths: Worked Examples Explained | A-Level 数学：典型例题详解